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Show that if $R$ is an integral domain then $R[X]^\ast=R^\ast$. Show that this is false if $R$ is not an integral domain.

Note1: this is not homework.Perhaps I must have mentioned that my biggest problem with attempting this question was the interpretation of $R[X]^\ast$ and $R^\ast$. It's for this reason that I noted this below yesterday.

Note2: this question actually comes from the notes of Iain Gordon(http://www.maths.ed.ac.uk/~igordon/).

$R^*$ here means the group of units of the ring $R$.

Jyrki Lahtonen
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    To prove that $R[X]^\subset R^$: if $p(x),q(x)\in R[x]$, then what is the degree of $p(x)q(x)$? When is this degree 0? – Andrea Feb 28 '16 at 22:16
  • One of my problems is that I don't understand how we can say that $R[X]^\ast=R^\ast$. On the left hand side equation we have a set of polynomials and on the right hand side we have an integral domain. How does equality make sense in the question? –  Feb 28 '16 at 22:20
  • If $;R;$ is an integral domain then $;r\to r+0\cdot x;$ is a ring embedding and thus we can consider $;R\subset R[x];$ . – DonAntonio Feb 28 '16 at 22:21
  • You can see each element $r\in R$ as a constant polynomial $r+0X+0X^2+...$. In other words, you have a natural map $R\to R[X]$ that identifies $R$ with a subring of $R[X]$. – Andrea Feb 28 '16 at 22:23
  • All: Comments are not for extended discussion; this conversation has been moved to chat. Unclear points were raised, and the comment exchange allowed the question to evolve into a better form. Thank you for your participation in improving the question – Jyrki Lahtonen Mar 02 '16 at 08:16
  • (cont'd) To repeat: The question has been edited. Please take that into account when judging the usefulness of answers. I suspect the current version may be a duplicate, so if you find a good duplicate target, please act accordingly. – Jyrki Lahtonen Mar 02 '16 at 08:28

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Suppose $f(x)\in R[x]$ is invertible; write it as $f(x)=a+xg(x)$ and suppose $b+xh(x)$ is the inverse. Then $$ ab+x(ah(x)+bg(x)+xg(x)h(x))=1 $$ and, comparing alike terms, we get $ab=1$. Therefore $a$ is invertible.

Until now we have not used the fact that $R$ is a domain. Suppose it is and that $f(x)$ has positive degree. Then $f(x)k(x)$ has positive degree: indeed the leading coefficient of the product is the product of the leading coefficients, which cannot be zero. Thus an invertible polynomial must have degree $0$ and, by what we saw before, it is an invertible constant.

Note that we are identifying constant polynomials in $R[x]$ with $R$.

In order to find a counterexample, we look at the simplest nondomain, that is, $R=\mathbb{Z}/4\mathbb{Z}$. The first case we can try is $1+2x$: the constant term must be invertible, and the leading coefficient must be a zero divisor, by the arguments above.

Now $$ (1+2x)^2=… $$

The general result that, if $R$ is not a domain, then the set of invertible polynomials contains non constant polynomials is false. See https://math.stackexchange.com/a/30390/62967, where it is proved that a polynomial $a_0+a_1x+\dots+a_nx^n$ is invertible if and only if $a_0$ is invertible and $a_i$ is nilpotent for $i>0$.

So if $R$ is not a domain but it has no (nontrivial) nilpotent element, then the invertible polynomials are just the invertible constants. A ring with such a property is the product of two fields, for instance.

Community
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egreg
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  • I think the question is to show that equality doesn't exist for any $;R;$ that is not an integral domain. If the ring has nilpotent elements then it is easy, but if not it looks like it is harder. – DonAntonio Feb 28 '16 at 22:23
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    @Joanpemo It is false, if $R$ has no nilpotent element: http://math.stackexchange.com/a/30390/62967 – egreg Feb 28 '16 at 22:29
  • Thank you very much. I thought so the question is either wrong or ill posed. – DonAntonio Feb 28 '16 at 22:30
  • @egreg: I don't see anything in the link in your comment that implies that $R$ must have nilpotent elements if $R[X]^* \neq R^*$. – Rob Arthan Feb 28 '16 at 22:54
  • @egreg I may be missing your point, but if $R^\ast$ is the group of units of the ring $R$ then it doesn't have a nilpotent element. So what's the issue? –  Feb 28 '16 at 22:56
  • @RobArthan Theorem. $a_0+a_1x+\dots+a_nx^n$ is invertible if and only if $a_0$ is invertible and each $a_i$, $i>0$, is nilpotent. – egreg Feb 28 '16 at 23:02
  • @AidanRocke If $R$ has no (nontrivial) nilpotent elements, then a polynomial is invertible if and only if it is an invertible constant. – egreg Feb 28 '16 at 23:04
  • @egreg: I agree, but see my answer. $R^* = R[X]^*$ implies that $R$ has no non-trivial nilpotent elements, but it doesn't imply that $R$ has no zero divisors. – Rob Arthan Feb 28 '16 at 23:05
  • @RobArthan I'm a bit confused. The definition of an integral domain is that $ab=0 \rightarrow a=0$ or $b=0$ but a nilpotent element $B$ has the property that $B^k = 0$ for some positive integer $k$. If we set $a=B^{k-1}$ and $b=B$ I think we have a contradiction. –  Feb 28 '16 at 23:08
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    @AidanRocke In a domain there's no nontrivial nilpotent element; the discussion is about the truth of $R[x]^\ne R^$ when $R$ is not a domain. – egreg Feb 28 '16 at 23:13
  • @egreg Ok. Now, I see what you mean.. –  Feb 28 '16 at 23:16
  • Your answer does not answer the second part of the question. – Rob Arthan Feb 28 '16 at 23:22
  • @RobArthan Sorry? What do you mean? The statement $R[x]^=R^$ can be true even if $R$ is not a domain; and it was in comments before you answered. Consider a product of two fields, which has no nontrivial nilpotent elements: the invertible polynomials are constant. I added the simple example, if that's what you mean. – egreg Feb 28 '16 at 23:26
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    Exactly: your answer did not address that part of the question. Thanks for ignoring my answer to that part of the question (which was intended to complement your excellent answer to the first part) and then (apparently) parroting it back to me. Cheers! – Rob Arthan Feb 28 '16 at 23:33
  • @RobArthan I find that your sarcasm is out of place. Joanpemo noticed that the question didn't ask for a counterexample, so I did some research and completed my answer; I wasn't aware of your answer, written after my reply to Joanpemo, from which a full argument can be easily derived. No “apparent parroting”, just the time needed for completing my answer. – egreg Feb 29 '16 at 09:25
  • @egreg: my sarcasm was intended to be very light-hearted, please accept my apologies for any offence. The situation was made worse because the question was changed several times: the original question amounted to a straightforward "iff" (which the second part of your answer did not cover) but it was changed later. – Rob Arthan Feb 29 '16 at 20:47
  • @RobArthan Maybe you took it too light-hearted. Your comment about me not having answered the second part of the question came after I edited it in. And I strongly suppose the downvote is yours. – egreg Feb 29 '16 at 22:41
  • @RobArthan I think there's a small misunderstanding here but I hope you will stop arguing with egreg. You both made your contributions independently from what I understand. But, my bigger issue is why is the question on hold all of a sudden? And, my question got edited somehow!? Who has been editing my question?? –  Feb 29 '16 at 22:49
  • @egreg: It takes me time to type too! It was the part of your answer beginning "in order to find a counter-example" that got me involved, because it really did not fit the right-to-left direction of the original statement of the question. – Rob Arthan Feb 29 '16 at 22:50
  • @AidanRocke: you can view the editing history by clicking on "edited 15 minutes ago" (or whatever it now says) under your question. I suspect the close votes that got your question put on hold were from people (not me) who thought you had not given enough information on your attempts to solve the problem. – Rob Arthan Feb 29 '16 at 22:58
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egreg's argument above shows that any unit in $R[x]$ is also a unit in $R$. If $R$ has no multiplicative identity, then neither does $R[x]$.

Since the degree-polynomials form a subring of $R[x]$ that is isomorphic to $R$, then $R[x]$ is commutative if and only if $R$ is commutative. Similarly, if $a, b \in R$ are non-trivial divisors of 0, then $a, b \in R[x]$ are likewise non-trivial divisors of 0.

Summarizing, if $R$ lacks any of the properties that distinguish a ring from an integral divisor, then $R[x]$ also lacks that property.

William Krebs
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  • I wonder what has this answer to do with the question? If $R$ has no unit, then how could we talk about invertible elements? – user26857 Feb 29 '16 at 00:21
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See egreg's answer to see why $R^* = R[X]^*$ when $R$ is an integral domain. The converse direction is false: take $R = \Bbb{Z} \times \Bbb{Z}$, then $R$ is not an integral domain, but $R[X]^*$, the group of units of $R[X]$, is the same as, $R^*$, the group of units of $R$.

Rob Arthan
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    I am fascinated to know the reason for that downvote. Sheer petty-mindedness, I imagine, given that the answer is a technically correct answer to part of the original question. – Rob Arthan Mar 02 '16 at 00:55
  • If a question suffers a radical edit, then I suppose we have to be wise enough to forget about the answers which don't match the actual question and simply delete them. Moreover, a question like "Show that this is false if R is not an integral domain" is clearly a logical mistake: it should be "Show that this can be false if R is not an integral domain." I've never seen someone asking for a counterexample by claiming that an assertion is false for every object which doesn't satisfy the hypothesis. This is why I think your answer really deserves a downvote. – user26857 Mar 13 '16 at 23:23